Bài 25:
a: \(\left(x-15\right)\cdot15=0\)
=>x-15=0
=>x=15
b: \(32\left(x-10\right)=32\)
=>\(x-10=\dfrac{32}{32}=1\)
=>x=1+10=11
c: (x-5)(x-7)=0
=>\(\left[{}\begin{matrix}x-5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)
d: 35(x-35)=35
=>x-35=35:35=1
=>x=1+35=36
Bài 26:
a: (x-15)-75=0
=>x-15=75
=>x=75+15=90
b: 575-(6x+70)=445
=>6x+70=575-445=130
=>6x=60
=>\(x=\dfrac{60}{6}=10\)
c: x-105:21=15
=>x-5=15
=>x=15+5=20
d: \(\left(x-105\right):21=15\)
=>\(x-105=21\cdot15=315\)
=>x=315+105=420
Bài 25:
\(a,\left(x-15\right).15=0\)
\(x-15=0:15\)
\(x-15=0\)
\(x=0+15\)
\(x=15\)
Vậy \(x=15\)
\(b,32.\left(x-10\right)=32\)
\(x-10=32:32\)
\(x-10=1\)
\(x=1+10\)
\(x=11\)
Vậy \(x=11\)
\(c,\left(x-5\right).\left(x-7\right)=0\)
\(x-5=0\) hoặc \(x-7=0\)
\(x=0+5\) hoặc \(x=0+7\)
\(x=5\) hoặc \(x=7\)
Vậy \(x\in\left\{5;7\right\}\)
\(d,\left(x-35\right).35=35\)
\(x-35=35:35\)
\(x-35=1\)
\(x=1+35\)
\(x=36\)
Vậy \(x=36\)
Bài 26:
\(a,\left(x-15\right)-75=0\)
\(x-15=0+75\)
\(x-15=75\)
\(x=75+15\)
\(x=90\)
Vậy \(x=90\)
\(b,575-\left(6x+70\right)=445\)
\(6x+70=575-445\)
\(6x+70=130\)
\(6x=130-70\)
\(6x=60\)
\(x=60:6\)
\(x=10\)
\(c,x-105:21=15\)
\(x-5=15\)
\(x=15+5\)
\(x=20\)
Vậy \(x=20\)
\(d,\left(x-105\right):21=15\)
\(x-105=15.21\)
\(x-105=315\)
\(x=315+105\)
\(x=420\)
Vậy \(x=420\)